Dear students, welcome to today's lecture. So I hope you have gotten the opportunity to go through your concepts of probability that we discussed in last lecture. So today we will continue our discussion on probability. So we will begin with a brief recap of what we had discussed in last class.
So we were discussing the concept of probability. So in probability, There are two things, one is a sample and one is an event. So of these the sample is defined as the list of all possible outcomes. And of these event will depend on how you define the event.
So in the general case number of possibilities in an event is less or equal to number of possibilities in the total sample space. So we discussed few cases so in case of a coin toss so for a coin toss experiment my sample space is h comma t. Now, this is the sample space for a single coin toss. So, this is one time.
So, if I have two successive coin tosses, So, my sample space becomes, so in the first case let us say first I get a head, I can get the second time also I can hit the head, this is one possibility. I can get first time head and second time tail, I can get the event the. first time is tail, the second time is a head or I can get two tails. So, for a single coin toss I have two possibilities, for a two time coin toss I have four possibilities. So, depending so as you.
increase the number of coin tosses you get more and more you know number of possibilities in your sample space ok. So, in this case so I can define an event as a head in this case ok. So, clearly you see so this is satisfied.
So, N of E in this case is less than number in sample space ok. So, there are three axioms first is probability of an event. event E is bounded between 0 and 1. So, if I defined an E which is so if my event E was neither a head nor a tail occurs appears. This event does not have any entity. So this is nothing but a null event.
So null is represent by the number 5. So in this case for a point or single point or six it is impossible that you will have neither a head nor a tail because the head and tail are the two of them. In this case my probability of E is going to return me a value of 0. So if I change my definition of E to the E as either a head or a tail appears. So, in this case my E is nothing but the sample space itself right.
So, you are happy with either a head or a tail ok. So, in this case the event is simply the sample size. So, in this case my probability of E is equal to 1 ok, but for the case of head appears E is head appears.
probability of E is going to be half. So these events can be you know Venn diagrams can be conveniently used to describe events and sample space. So sample space is typically drawn as a box and events are drawn as circles. So this is an event E1.
is an event E2, let us say this is an event E3. So, I can write down we have derived these expressions probability of E1 union E2. So, this is union is equal to pe1 plus pe2 minus p of E1. Now many places instead of writing like this you can also you will see E1 plus E2 will be written as pe1 plus pe2 minus probability of E1 E2, okay.
So this is an intersection. So, many places instead of a union you will see a plus input and instead of E1 intersection E2 you will get the expression as E1 into E2. So, what I have also drawn here I have drawn an event E3 which does not overlap with either E1 or E2. So, these are examples of mutually exclusive events.
So, I can write P of E1 E2 intersection E2 or P of E1 E2 is equal to 0 because there is nothing in commonality between the two events P E1 E2 is 5. So I can for any two events so for mutually exclusive events. Probability of union of Ei, i equal to 1 to n is nothing but equal to 1 and this is equal to summation probability of Ei. Because each of the intersections, so in this case in the most generic case for probability of EI intersection EJ is going to be 0. So in every addition is simply these individual elements which are getting added up.
So this is the second thing. Now let us take an example. Let us have consider a case, we have a box which has 5 blue balls and 3 red balls.
So, you have 5 blue and 3 red balls. So, this is a container having 5 blue and 3 red. So, I can ask a question what is the chance probability of this?
drawing two red and one blue ball from this box. So, what is the probability of drawing 2 red and 1 blue ball from this box? So, how do I do it?
So, let us just say if I am collecting at 1, I am basically taking out 3 balls. So, I have to take out 3 balls. I have a total of 5 plus 3 equal to 8 balls. So, in order to calculate to answer this question we need to come up with ways of counting.
So that brings us to two ideas important in the process of permutations and combinations. To answer this question I need to briefly discuss permutations and combinations. So, imagine let us take a very simple case.
I have these three numbers a, b, c and let us think about in what ways can I arrange a, b, c in a linear order. So, I can write them as so ordering of a, b, c. So, I can write them as ABC, I can write them as ACB, I can write them as BCA, I can write it as BAC, I can write it as CAB and CBA. So, there are six possibilities. So, the way to find out, find this out is ask, so you have three positions and you have three numbers a, b and c.
You want to put them here. Now, imagine in how many ways, so imagine you have three slots which were empty. In how many ways can you fill up the first three slots, the first slot? So, I can either put A, I can either put or put B or put C.
So, there are three ways of putting. or filling up the first slot. So, I have this condition, I have three ways of filling up the first slot. After I filled up the first slot, let us just hypothetically take an example, my first slot is A. In how many ways can I fill up the next slot?
So, I am left once since I have put A here, I am left with B and C only. So, here I can either put B or I can put C. So, there are two possibilities here.
And once I have put b here, I am left with no alternative but only I have to put c here because that is the only letter which is still available to me or vice versa. Similarly, if I put c here, the only other way is to put b here. So, for each of these three ways, then there are two ways.
So, the way to do it, so this is basically for the first step into 2 ways for the second step into 1 way for the first third step. So, this is how you get a total of 6 possibilities. In the general case if there are n 1 ways n 1 ways of putting in the first slot I write down the expression n 1 then you have n 2 n 2 ways of fitting up the second slot.
So, this is in the general case when you have n objects and then you want to find out r objects from that. So, you want to order take r objects from n objects and order them. So, I can there might be n 1 times n 1 possibilities for the first object, n 2 possibilities for the first object, second object, second slot and so on and so forth.
So, if there are n objects then I know n 1 is nothing but is equal to n. So, I can do for n objects the first slot can be for first slot there are n ways. So, for n objects. Second slot there are n minus 1 ways.
Third slot n minus 2 is. So, you see it is decreasing by 1. So, the total number of possibilities is nothing but n into n minus 1 into n minus 2 up to 1. This number is nothing but the factorial of the number n. This is you write the fact. So, this is the factorial.
of the number n. This is also or n exclamation is also used for describing the factorial of the number n. So factorial of 1 is simply is equal to 1, factorial of 2 is equal to 1 into 2. So factorial of 0 is also defined as 1, factorial of 3 is 3 into 2 into 1 is equal to 6. 4 x 3 x 2 x 1 is 24 so on and so forth.
So there are n factorial ways of arranging n objects in a linear order. So imagine you have a bookcase, imagine consider the case you have 4 maths, 4 biology, 4 math, and two English books and you want to arrange them in a book case such that books of the same subject. are next to each other.
So how will I do it? So let's say I name my maths books. So I have three categories. Categories are maths, biology and English. This is for maths, this is for biology.
and this is for English, Maths, Biology and English. And in Maths you have four books, you can label them as M1, M2, M3 and M4. In Biology you have B1, B2, B3 and B4. And you have English. E1 and E2 ok.
How will you arrange? So, there are three categories right. First is you think about it if this is your book rack ok, this is your book rack. I can have of three categories ok, category 1, category 2, category 3. So, in three categories there are three categories of books Maths, Biology and English.
I can order them three categories in how many different ways we just saw which is going to be factorial 3 is equal to 6 possible ways. Why because what are the six combination I could have put MBE, I could have put Maths, English, Biology, I could have put BME, Maths, I could have put EMB EBM. So you have 6 combinations which is factorial 3 and 3 categories you can be arranged in a book rack in 3 factorial ways. Now for each of these categories, so imagine you have a one particular setup which is MBE. So one particular setup you have chosen let us say MBE.
Within this M you actually have 4 combinations. You have to arrange M1, M2, M3, M4. Similarly within the B you have to arrange B1, B2, B3, B4 and within the E you have to arrange E1 and E2. So, 4 sets of books. Within a block can be arranged in factorial four ways.
Why? Because the first slot, so again you have four slots, the first slot can be filled it in four different ways. Once you have put a given number, let us just say m1 you have put here, then in the second slot there are one of three different numbers can be chosen here. So, you can have three ways, then the next one is two ways, and the next one is one way. So, factorial four ways is a number of arranging here.
Similarly, factorial 4 ways is the number of possible arrangements here and factorial of 2 is the arrangement here. So, the total number of possible arrangements comes out. So, this is once you have decided on the category 4 factorial into 4 factorial into 2 factorial. So, these are the number of possibilities you see there are products of 4 into 4 into 2 is the number of categories and for each category for a given choose of a category order.
Within this you have 4 factorial for each of this ordering in maths you can have another 4 factorial within biology which is why you multiply by 4 and this is by 2. And the final answer is why you multiply it by factorial 3 because these are the categories. So, category ordering. So, thus your answer so your sample space has factorial 3 into 4 into 4 into 3 which is. 6 into 24 into 24 into 2. So, 24 square is 576. So, you have 6 into 2 into 576. You can calculate the final. So, you have these many number of possibilities.
So, this represents, so n of sample space represents 6 into 2 into 576. Now, let us define the event E as the ordering such that maths is at the beginning. that is on the left. So, this is the possibility of event of E.
So, what are the categories that are possible? The number of categories which are possible are basically you have MBE or MEB. But, for each of these two categories for each of so I have so the total number of possibilities of N of E is going to be 2 times the remaining thing is as before ok. Within each of these orderings you can again rearrange as per the maths or the bio the English as before.
So, if N of E is simply 2 into 4 factorial 4 factorial into 2 factorial ok. So, then your probability of the event E is nothing by 2 into factorial 4, factorial 4, factorial 2. by 6 into factorial 4, factorial 4, factorial 2 is nothing by 2 by 6 is equal to one third. So this is how you order.
There is a slightly different thing from per so this is ordering is called permutation or arrangement ok. Now imagine you want to you have 10 students in a class and you want to constitute a committee. So of these 10, let us assume that there are 4 girls and 6 boys.
You have 4 girls in the class and 6 boys in the class. And you want to constitute a committee of 3 members. Okay, such that there are boys such that there are two girls and one boys.
Okay, so in a committee when you constitute a committee what matters. is not the order. So, let us say I have a committee I have let us say g1, g2, b1.
So, my girls are g1, g2, g3, g4 and my boys are b1, b2 up to b6. So, in one particular set let us say I have committee 1 is g1 g2 b1, committee 2 is g2 g1 b1 ok. See the ordering is different, but for a committee the order does not matter.
So this is the same event ok, I have the same event. So the number of possibilities ok, so here you are not concerned about. but you are considering of the selection. So, in general the ways the number of ways of selecting r from a population of n is represented. by n C R and n C R is defined by factorial n by factorial R, factorial of n minus R.
So this is different. So this is defined as n P R. So, this is permutation, permutation NPR is defined by factorial n by factorial n minus r and NCR is defined by a factorial n factorial r by factorial n minus r. So, you see there is this division by this factorial r. So, this is an example. So, the ordering can be more, but in terms of combination it is the still the same combination ok.
So, that is why you divide by factorial r. So, nPr is factorial n by factorial of n minus r, nCr is equal to factorial n by factorial r, factorial n minus r. So, n's, so let's say if n is 3, 3C1 is going to be factorial 3 by factorial 1, factorial 2 is simply equal to 3. versus 3 p 1 is equal to factorial 3 by factorial 0 is equal to 3. 3 c 2 is same as, so you see the definition of n c r if I put instead of r if I put n minus r if I put n minus r by r you get the same thing. So, n c r is same as n c n minus r.
So, as is the case of 3 c 1 and 3 c 2. So, factorial 2 and factorial 1 you still get the value of 3. So, n c r is same as n c n minus r. So, the way of choosing let us say you have 10 students where you have 4 girls and 6 boys and you want to constitute a committee of 2 girls and 1 boy. So the number of possibilities so you want to choose 4 out of 10 you want to do 4C2 and for each you want to do 6C1. This is your n of sample space. 4C2 into 6C1 is the total number of ways in which you can constitute a committee of two girls and one boy from this.
Now, so this is event E1. So, I can say the event is such that this, but in how many different ways can you choose 3 out of 10. So, your n of sample space is 10C3. So, 10C3.
So, I can ask the question what is the chance that you have come up with a committee which has 2 girls and 1 boy ok. So, the answer is then so probability of constitutive committee which has 2 girls and 1 boy is simply 4 c2 into 6 c1 by 10 c3 ok. So, this is your probability of this event. is the event is the possibility that you have constituted a committee of two girls and one boy ok.
Let us take another example. So let us say you have n objects from which you want to draw k objects ok, but you always so let us say there is an element a which you always want to select. So, what is the probability that you choose k out of n such that a is always selected? So, to answer this question my probability ok.
So, in how many ways can I choose k from n? The answer is n c k. So, this is the probability of your sample associated with your sample n of s ok.
So, if there is a element a which is always selected the probability of selecting k out of n is equivalent of saying that I want to select k minus 1 from n minus 1 ok. So, which is nothing but n minus 1 c k minus 1 ok. So, probability of selecting this a say that a is always selected is nothing but n minus 1 ck minus 1 by n ck.
So, with that I stop for today. Just to briefly summarize we discussed about. the basics of probability once more and we chose.
So, one of the critical steps of calculating probability is to do a counting ok. And then there are two ways of counting which we call a permutation or a combination. A permutation is done when you have a ordering or an arrangement. A combination is done when you just want to select and the order does not make a may have any importance as in case of constituting a committee or something like that. With that I thank you for your attention and I look forward for our next lecture.